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Acute and obtuse angles

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Say True or False.
The measure of a reflex angle $> 180^o$.

  1. True

  2. False


Correct Option: A
Explanation:
According to defition , reflex angle is the angle which are greater than $180^o$ and less than $360^o$.
So,the statement is true

An angle which is more than $180^{\circ}$ and less than $360^{\circ}$ is called:

  1. obtuse angle

  2. right angle

  3. reflex angle

  4. complete angle


Correct Option: C
Explanation:

An angle which is more than 180180∘ and less than 360360∘ is called reflex angle.

So option $C$ is correct.

An angle which is more than  $\displaystyle 180^{0}$ and less than $\displaystyle 360^{0}$ is called

  1. Obtuse angle

  2. right angle

  3. reflex angle

  4. complete angle


Correct Option: C
Explanation:

An angle which is more than  $\displaystyle 180^{0}$ and less than $\displaystyle 360^{0}$ is called reflex angle.


If one angle at a point is reflex angle, the other at that point may be :

  1. Acute angle

  2. Obtuse angle

  3. Straight angle

  4. Acute or obtuse angle


Correct Option: D
Explanation:

Angle formed at a point is $\displaystyle { 360 }^{ o }$. 

When one of angle is reflex it range from $\displaystyle { 180 }^{ o }$ to $\displaystyle { 360 }^{ o }$. 
If angle is $\displaystyle { 200 }^{ o }$ then the other angle is $\displaystyle { 160 }^{ o }$, i.e obtuse angle. 
If one reflex angle is $\displaystyle { 300 }^{ o }$ then other angle is $\displaystyle { 60 }^{ o }$ i.e acute angle.

Sum of two obtuse angle results in:

  1. Acute angle

  2. Right angle

  3. Obtuse angle

  4. Reflex angle


Correct Option: D
Explanation:

Obtuse angle ranges from $\displaystyle { 90 }^{ o }$ to $\displaystyle { 180 }^{ o }$
So, sum of least obtuse angle is $\displaystyle { 91 }^{ o }+{ 91 }^{ o }={ 182 }^{ o }$ which is a reflex angle.
Sum of maximum obtuse angle is $\displaystyle { 179 }^{ o }+{ 179 }^{ o }={ 358 }^{ o }$ which is a reflex angle.

Acute angle is the angle which is greater than $\displaystyle { 0 }^{ o }$ and less than $\displaystyle { 90 }^{ o }$.

Obtuse angle is the angle which is greater than $\displaystyle { 90}^{ o }$ and less than $\displaystyle {180 }^{ o }$.
Right angle is the angle which is equal to $90^o$.
Reflex angle is the angle which is greater than $\displaystyle {180}^{ o }$ and less than $\displaystyle {360 }^{ o }$.

Which of the following is a reflex angle?

  1. $\displaystyle { 180 }^{ o }$

  2. $\displaystyle { 360 }^{ o }$

  3. $\displaystyle { 204 }^{ o }$

  4. $\displaystyle { 135 }^{ o }$


Correct Option: C
Explanation:

Reflex angle must be between $\displaystyle { 180 }^{ o }$ and $\displaystyle { 360 }^{ o }$. It can't be equal to $\displaystyle { 180 }^{ o }$ or $\displaystyle { 360 }^{ o }$ as these are straight and complete angle respectively.

Acute angle is the angle which is greater than $\displaystyle { 0 }^{ o }$ and less than $\displaystyle { 90 }^{ o }$.

Obtuse angle is the angle which is greater than $\displaystyle { 90}^{ o }$ and less than $\displaystyle {180 }^{ o }$.
Right angle is the angle which is equal to $90^o$.
Reflex angle is the angle which is greater than $\displaystyle {180}^{ o }$ and less than $\displaystyle {360 }^{ o }$.

Mark the correct alternative of the following.
A reflex angle measures.

  1. More than $90^o$ but less than $180^o$

  2. More than $180^o$ but less than $270^o$

  3. More than $180^o$ but less than $360^o$

  4. None of these


Correct Option: C
Explanation:

An angle is said to be a reflex angle if it measures more than $180^o$ but less that $360^o$.

Let the slope of the lines upon which the incident ray and line mirror lie are respectively $5$ and $3$, then the slope of the line upon which the reflected ray lies is 

  1. $1$

  2. $2$

  3. less than $2$

  4. more than $2$


Correct Option: A

If the angle of a triangle are in the ratio of $2:3:4$, find the there angles 

  1. $40^{o}, 60^{o}, 80^{o}$

  2. $20^{o}, 40^{o}, 60^{o}$

  3. $30^{o}, 60^{o}, 90^{o}$

  4. $90^{o}, 180^{o}, 360^{o}$


Correct Option: A
Explanation:

Let the angles of the triangle be $2x,3x, 4x$. 

Then
$2x+3x+4x=180^o$ [ Since sum of the interior angles of a triangle be $180^o$]
or, $9x=180^o$
or, $x=20^o$.
Then the angles of the triangle be $40^o, 60^o, 80^o$.

Say True or False.
The measure of an acute angle $< 90^o$.

  1. True

  2. False


Correct Option: A
Explanation:
By definition acute angle is the angle which are less than $90^o$.
So, the above statement is true

Say True or False.
If $m\angle A=53^o$ and $m\angle B=35^o$, then $m\angle A > m\angle B$.

  1. True

  2. False


Correct Option: A
Explanation:
$True$
$m \angle A=53^ \circ  \ and \ m \angle B=35^ \circ$
$\because 53 \ > \ 35$
$\therefore m \angle A> m \angle B$

An angle which measures $0^{\circ}$ is called

  1. Obtuse

  2. Straight

  3. Zero

  4. Right


Correct Option: C
Explanation:

An angle which measures $ {0}^{o} $ is called zero angle.

A, B, C and D are four angles at a point so that $A+B+C+D=4$ rightangles, outof these A and B are acute angles while C and D are obtuse angles. Which of the following relations may be true?

  1. $A+B=C+D$
  2. $A+C=B+D$
  3. $A+D=B+C$

  1. 2 and 3 only

  2. 1 and 3 only

  3. 1 and 2 only

  4. 3 only


Correct Option: A

An angle which measures $\displaystyle 0^{0}$ is called-

  1. Zero

  2. Obtuse

  3. Right

  4. None of these


Correct Option: A
Explanation:

an angle which measures 0 degree is called a zero angle..

In a $\displaystyle \Delta PQR$ PQ = PR and $\displaystyle \angle Q$ is twice that of $\displaystyle \angle P$ Then $\displaystyle \angle Q$__

  1. 75$\displaystyle ^{\circ}$

  2. 65$\displaystyle ^{\circ}$

  3. 72$\displaystyle ^{\circ}$

  4. 100$\displaystyle ^{\circ}$


Correct Option: C
Explanation:

$\displaystyle \because $ PQ = PR 
$\displaystyle \Rightarrow $ $\displaystyle \angle Q=\angle R$
Given that $\displaystyle \angle Q=2\angle P$
We have
$\displaystyle \angle P+\angle Q+\angle R=180^{\circ}$
$\displaystyle \frac{\angle Q}{2}+\angle Q+\angle Q=180^{\circ}$
$\displaystyle \frac{5}{2}\angle Q=180^{\circ}$
$\displaystyle \angle Q=72^{\circ}$        

Find the angle between the lines 3x + 2y = 6 and x + y = 6

  1. 12$\displaystyle ^{\circ}$ 20'

  2. 11$\displaystyle ^{\circ}$ 19'

  3. 14$\displaystyle ^{\circ}$ 25'

  4. 13$\displaystyle ^{\circ}$ 06'


Correct Option: B
Explanation:

Line  $3x+2y=6$

$=>2y=-3x+6$
$=>y=\frac{-3}{2}x+\frac{6}{2}$
$=>m'=\frac{-3}{2}x$
$\therefore tan\theta _1=\frac{-3}{2}$
$=>\theta _1=tan^{-1}(\frac{-3}{2})$
$=>\theta _1=-56.31$
Line is $x+y=6$
$=>y=x+6$
$=>m _2=-1$
$=>tan\theta _2=-1$
$=>\theta _2=tan^{-1}(-1)$
$=>\theta _2=-45$
Thus, ale between them $=(-45-56.31)$
                                          $=11.31^0$

Find the complement of each of the following angles $24^{\circ}$

  1. $66^{\circ}$

  2. $156^{\circ}$

  3. $36^{\circ}$

  4. None of these


Correct Option: A
Explanation:

We know that the complement angle

$=90^0-\theta$

So,
The complement angle of $24^0$ will be
$=90^0-24^0=66^0$ 

Hence, this is the answer.

Find the complement of each of the following angles
$63^{\circ}$

  1. $27^{\circ}$

  2. $54^{\circ}$

  3. $117^{\circ}$

  4. None of these


Correct Option: A
Explanation:

We know that the complement angle

$=90^0-\theta$

So,
The complement angle of $63^0$ will be
$=90^0-63^0=27^0$ 

Hence, this is the answer.

Find the angles in each of the following.
The angle whose complement is one sixth of its supplement

  1. $72^{\circ}$

  2. $32^{\circ}$

  3. $62^{\circ}$

  4. None of these


Correct Option: A
Explanation:

Let the angle be $x^0$.


Supplement angle $=(180^0-x)$

Complement angle $=(90^0-x)$

According to the question,
$\dfrac{1}{6}\times (180^0-x)=(90^0-x)$

$(180^0-x)=6(90^0-x)$

$180^0-x=540^0-6x$

$5x=360^0$

$x=72^0$


Hence, this is the answer.

Find the angles in each of the following.
The angle which is four times its supplement

  1. $144^{\circ}$

  2. $44^{\circ}$

  3. $14^{\circ}$

  4. None of these


Correct Option: A
Explanation:

Let the angle be $x^0$.


According to the question,
Supplement angle $=(180^0-x)\times 4$

So,

$x=720^0-4x$

$5x=720^0$

$x=144^0$


Hence, this is the answer.

Find the angles in each of the following.
The angles whose supplement is four times its complement

  1. $60^{\circ}$

  2. $30^{\circ}$

  3. $120^{\circ}$

  4. None of these


Correct Option: A
Explanation:

Let the angle be $x^0$.


Supplement angle $=(180^0-x)$

Complement angle $=(90^0-x)$

According to the question,
$180^0-x=4(90^0-x)$

$180^0-x=360^0-4x$

$3x=180^0$

$x=60^0$


Hence, this is the answer.

Find the supplement of each of the following angles.
$148^{\circ}$

  1. $32^{\circ}$

  2. $122^{\circ}$

  3. $148^{\circ}$

  4. None of these


Correct Option: A
Explanation:

We know that the supplement angle

$=180^0-\theta$

So,
The supplement angle of $148^0$ will be
$=180^0-148^0=32^0$ 

Hence, this is the answer.

Find the supplement of each of the following angles.
$120^{\circ}$

  1. $60^{\circ}$

  2. $150^{\circ}$

  3. $20^{\circ}$

  4. None of these


Correct Option: A
Explanation:

We know that the supplement angle

$=180^0-\theta$

So,
The supplement angle of $120^0$ will be
$=180^0-120^0=60^0$ 

Hence, this is the answer.

Find the complement of each of the following angles $35^{\circ}$

  1. $55^{\circ}$

  2. $145^{\circ}$

  3. $35^{\circ}$

  4. None of these


Correct Option: A
Explanation:

We know that the complement angle

$=90^0-\theta$

So,
The complement angle of $35^0$ will be
$=90^0-35^0=55^0$ 

Hence, this is the answer.

Find the supplement of the given angle.
$100^{\circ}$

  1. $80^{\circ}$

  2. $40^{\circ}$

  3. $30^{\circ}$

  4. None of these


Correct Option: A
Explanation:

We know that the sum of the supplement angles

$=180^0$

So,
The supplement angle of $100^0$ will be
$=180^0-100^0=80^0$ 

Hence, this is the answer.

Find the complement of each of the following angles $20^{\circ}$

  1. $70^{\circ}$

  2. $60^{\circ}$

  3. $110^{\circ}$

  4. None of these


Correct Option: A
Explanation:

We know that the complement angle

$=90^0-\theta$

So,
The complement angle of $20^0$ will be
$=90^0-20^0=70^0$ 

Hence, this is the answer.

Find the angles in each of the following.
The angle which is two times its complement

  1. $60^{\circ}$

  2. $120^{\circ}$

  3. $30^{\circ}$

  4. None of these


Correct Option: A
Explanation:

Let the angle be $x^0$.


According to the question,
Complement angle $=(90^0-x)\times 2$

So,

$x=180^0-2x$

$3x=180^0$

$x=60^0$


Hence, this is the answer.

Find the complement of each of the following angles $48^{\circ}$

  1. $42^{\circ}$

  2. $52^{\circ}$

  3. $132^{\circ}$

  4. None of these


Correct Option: A
Explanation:

We know that the complement angle

$=90^0-\theta$

So,
The complement angle of $48^0$ will be
$=90^0-48^0=42^0$ 

Hence, this is the answer.

Find the angles in each of the following.
Two complementary angles are in the ratio $3 : 2$

  1. $54^{\circ}, 36^{\circ}$

  2. $44^{\circ}, 36^{\circ}$

  3. $54^{\circ}, 46^{\circ}$

  4. None of these


Correct Option: A
Explanation:

Let the angles be $3x$ and $2x$.


So,
$3x+2x=90^0$

$5x=90^0$

$x=18^0$

Therefore, the angle are
$54^0, 36^0$


Hence, this is the answer.

If $\angle A$ is complement to $30^o$ and $\angle B $ is supplement to $120^o$ then:

  1. $\angle A > \angle B$

  2. $\angle A < \angle B$

  3. Incomparable

  4. $\angle A = \angle B$


Correct Option: D
Explanation:

Sum of complementary angles is $90^{\circ}$

$\Rightarrow \angle A+{ 30 }^{ \circ  }={ 90 }^{ \circ  }\ \Rightarrow \angle A={ 60 }^{ \circ  }$
Sum of supplementary angles is $180^{\circ}$
$\Rightarrow \angle B+{ 120 }^{ \circ  }={ 180 }^{ \circ  }\ \Rightarrow \angle B={ 60 }^{ \circ  }$
$\Rightarrow \angle A=\angle B$

Which is the greatest angle in the given set: $\dfrac{1}{3}$ of complete angle, $\dfrac{1}{3}$ of straight angle or a right angle?

  1. $\dfrac{1}{3}$ of complete angle

  2. $\dfrac{1}{3}$ of straight angle

  3. A right angle

  4. All are equal


Correct Option: A
Explanation:

Complete angle $=360^{\circ}$

$\dfrac{1}{3}$ complete angle $=\dfrac { 1 }{ 3 } \times { 360 }^{ \circ  }={ 120 }^{ \circ  }$
Right angle $=90^{\circ}$
$\dfrac{1}{3}$ right angle $=\dfrac { 1 }{ 3 } \times { 90 }^{ \circ  }={ 30 }^{ \circ  }$
So $\dfrac{1}{3}$ complete angle is greater.

Rank the following angles in descending order. 
1. Straight angle
2. Reflex angle
3. Right angle 

  1. $2$, $1$, $3$

  2. $3$, $2$, $1$

  3. $2$, $3$, $1$

  4. $3$, $1$, $2$


Correct Option: A
Explanation:
  1. Right angle $=90^{\circ}$

    2. Straight angle $=180^{\circ}$

    3. Reflex angle lies between $180^{\circ}$ and $360^{\circ}$

    So the descending order of angles is
    $2>1>3$

In a triangle, the angles are in ratio $1: 3: 2$. Find the difference between the greatest and smallest angle of the triangle.

  1. $10^o$

  2. $70^o$

  3. $60^o$

  4. $20^o$


Correct Option: C
Explanation:

Le the angles be $x,3x$ and $2x$

Using angle sum property of triangle 
$x+3x+2x={ 180 }^{ \circ  }\ 6x={ 180 }^{ \circ  }\ \Rightarrow x={ 30 }^{ \circ  }$
So the angles are 
$x={ 30 }^{ \circ  }\ 3x=3\times { 30 }^{ \circ  }={ 90 }^{ \circ  }\ 2x=2\times { 30 }^{ \circ  }={ 60 }^{ \circ  }$
Difference between largest and smallest $={ 90 }^{ \circ  }-{ 30 }^{ \circ  }={ 60 }^{ \circ  }$

If the difference of two supplementary angles is $40^{\circ}$, then the measurement of the greater angle is

  1. $65^{\circ}$

  2. $110^{\circ}$

  3. $130^{\circ}$

  4. $220^{\circ}$


Correct Option: B
Explanation:

Let the two supplementary angles are $x^{\circ}$ and
$180^{\circ} - x$
By hypothesis, $x - (180^{\circ} - x) = 40^{\circ}$
or $2x - 180^{\circ} = 40^{\circ}$
or  $2x = 220^{\circ}$
or    $x = 110^{\circ}$

In a $\Delta$ PQR, if $3\sin P+4\cos Q=6$ and $4 \sin Q+3\cos P=1$, then the angle $R$ is equal to :

  1. $\dfrac{3\pi}{4}$

  2. $\dfrac{5\pi}{6}$

  3. $\dfrac{\pi}{6}$

  4. $\dfrac{\pi}{4}$


Correct Option: B
Explanation:

Given trignometric equations are:

$3 \sin{P} +4 \cos{Q} =6$ -------(1)
$4 \sin{Q} +3 \cos{P} =1$ -------(2)

Squaring both equations (1) and (2) and adding them, we get
$\Rightarrow 9\left( \sin ^{ 2 }{ P } +\cos ^{ 2 }{ P }  \right) +16\left( \sin ^{ 2 }{ Q } +\cos ^{ 2 }{ Q }  \right) +24\left( \sin { Q } \cos { P } +\cos { Q } \sin { P }  \right) =36+1$

$ \Rightarrow 9+16+24\sin { \left( P+Q \right)  } =37$

$ \therefore \sin { \left( P+Q \right)  } =\cfrac { 37-25 }{ 24 } =\cfrac { 12 }{ 24 } =\cfrac { 1 }{ 2 } $

$\therefore P+Q=30°$

Hence, angle $R=180°-30°=150°=\cfrac { 5\pi  }{ 6 } $radian

In triangle $ABC,$ if $\dfrac { 1 }{ a+c } +\dfrac { 1 }{ b+c } =\dfrac { 3 }{ a+b+c } ,$ then $\angle c$  is equal to:

  1. $30^{\circ}$

  2. $45^{\circ}$

  3. $60^{\circ}$

  4. $90^{\circ}$


Correct Option: C
Explanation:

Given: $\cfrac { 1 }{ a+c } +\cfrac { 1 }{ b+c } =\cfrac { 3 }{ a+b+c } $


$\Rightarrow \quad \cfrac { a+b+2c }{ (a+c)(b+c) } =\cfrac { 3 }{ a+b+c } $

$ \therefore (a+b+2c)(a+b+c)=3(a+c)(b+c)$

$\Rightarrow$ $ { a }^{ 2 }+ab+ac+ab+{ b }^{ 2 }+bc+2ac+2bc+2{ c }^{ 2 }$$ =3(ab+ac+bc+{ c }^{ 2 })$

$ \therefore { a }^{ 2 }+2ab+3ac+{ b }^{ 2 }+3bc+2{ c }^{ 2 }$$ =3ab+3ac+3bc+3{ c }^{ 2 }$

$\Rightarrow$ ${ a }^{ 2 }+{ b }^{ 2 }=ab+{ c }^{ 2 }$

$\Rightarrow$ $ { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 }=ab$

$\Rightarrow$ $ \cfrac { { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } }{ ab } =1$

$\Rightarrow \cfrac { { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } }{ 2ab } =\cfrac { 1 }{ 2 } $

$\cos { C } =\dfrac{1}{2}\Rightarrow \angle C={ 60 }^{\circ}$

In $\Delta ABC\,,\,if\,\,A\,\,:\,\,B\,:\,\,C\, = \,1\,\,:\,\,5\,\,:\,\,6\,\,then$ find the value of $\sin A: \sin B: \sin C$

  1. $\left( {\sqrt 3 \, - \,1} \right)\,:\,2\sqrt 2 \,:\,\left( {\sqrt 3 \, + \,1} \right)$

  2. $2\sqrt 2 \,:\,\left( {\sqrt 3 \, - \,1} \right)\,:\,\left( {\sqrt 3 \, + \,1} \right)$

  3. $ \,\left( {\sqrt 3 \, - \,1} \right)\,:\,\left( {\sqrt 3 \, + \,1} \right)\,:\,2\sqrt 2 $

  4. $ \,\left( {\sqrt 3 \, - \,1} \right)\,:\,\sqrt 3 :\,\sqrt 2 $


Correct Option: C
Explanation:

Given that $ A:B:C = 1:5:6 $


We know that the sum of angles in a triangle is $180^0$

Let us first find each angle.

Total no.of parts= $1+5+6= 12$

$A$= $ \dfrac{1}{12} $ ($180^0$) =$15^0$

$B$= $ \dfrac{5}{12} $ ($180^0$) =$75^0$

$C$= $ \dfrac{6}{12} $ ($180^0$) =$90^0$

Hence, $ \sin A: \sin B: \sin C$ = $\sin 15^o$$:\sin 75^o$$:\sin 90^o$

$=\dfrac {\sqrt{3}-1} {2\sqrt 2} : \ \dfrac{\sqrt3+1}{2\sqrt2 } :1$ 

$= ({\sqrt{3}-1})  : ({\sqrt3+1} ): ({2\sqrt2 }) $ 

In $\Delta ABC$ and $\Delta DEF$, we have $\dfrac {AB}{DE}=\dfrac {BC}{FD}$. Triangles ABC and DEF will be similar if :

  1. $\angle A = \angle D$

  2. $\angle A = \angle F$

  3. $\angle B = \angle E$

  4. $\angle B = \angle D$


Correct Option: D

Using ruler and compasses only, construct a triangle POR such that $\angle P = 120^{\circ}$, PO = 5 cm PR = 6 cm.In the same figure, find a point which is equidistant from its sides. Name this point With this point as centre draw a circle touching all the sides of the triangle.

  1. Circumcentre

  2. Incentre

  3. Mid point

  4. Data insufficient


Correct Option: B
Explanation:

(I) We draw a triangle POR with $\quad \angle RPO={ 120 }^{ O }.\quad $

(II) OI & OR are the angular bisectors of $\quad \angle RPO\quad &amp; \angle ROP\quad $
The bisectors intersect at I.
(i) IM & IN  are drawn perpendiculars from i to PR & PO respectively. 
 (iii) The circle which touches the sides of the triangle POR.
has the radius  IM=IN.
Justification-
Between $\quad \Delta IPM\quad &amp; \quad \Delta IPN,\ \angle IMP={ 90 }^{ o }=\angle INP,\ \angle IPM=\angle IPN\quad $
So the third angles  $\quad \angle PIN=\angle PIM\quad $
Also the side IP is common.
So, by ASA rule, $\quad \Delta IPM\equiv \Delta IPN.\quad $  
i.e IM=IN.
Similarly, by considering the triangles INO & ISO it can be shown that
IN=IS.
So IM=IN=IS.
i.e the circle with centre I touches the sides of the given triangle.
So  I is the INCENTRE of the triangle POR.
Ans- Option B.

The measure of the trisected angle of $\displaystyle 162^{\circ}$ is:

  1. $\displaystyle 54^{\circ}$

  2. $\displaystyle 81^{\circ}$

  3. $\displaystyle 162^{\circ}$

  4. $\displaystyle 90^{\circ}$


Correct Option: A
Explanation:

Trisected angle $\displaystyle =\frac{162^{\circ}}{3}=54^{\circ}$

The difference between two angles is $19$$\displaystyle ^{o}$ and their sum is $\displaystyle \frac{890}{9}^o$. Find the greater angle.

  1. $63^o$

  2. $35^o$

  3. $27^o$

  4. $59^o$


Correct Option: D
Explanation:
Let the two angles be $a$ and $b$

Then, 
$a  - b = 19$

and, $a + b = \dfrac{890}{9}$

Adding the two equations,

$2a = \dfrac{890 + 19\times 9}{9}$

$2a = \dfrac{1061}{9}$

Thus, $a = \dfrac{1061}{18}$

$a \approx 59^{\circ}$

If two angles of a triangle are acute angles, the third angle:

  1. is less than the sum of the two angles

  2. is an acute angle

  3. is the largest angle of the triangle

  4. may be an obtuse angle


Correct Option: D
Explanation:
For a triangle $ABC$, sum of angles is $180^0$
Two angles are given as acute, $\angle A, \angle B < 90^0$
$\angle A = 90^0 - x$
$\angle B = 90^0 - y$
where, $x,y < 90^0$
$\therefore \angle A + \angle B+ \angle C = 180^0$
$\Rightarrow \angle C = x+y $
$x+y$ can be $>90^0$ or $<90^0$
So, it may be obtuse or acute.
Hence, option D.

An angle which measures $\displaystyle 0^{o}$ is called:

  1. obtuse angle

  2. straight angle

  3. zero angle

  4. right angle


Correct Option: C
Explanation:

An angle which measure $0^0$ is called zero angle.

An angle which measure more than $0^o$ and less than $90^o$ is an acute angle.
An angle which measures $90^o$ is called a right angle.
An angle which measure more than $90^o$ and less than $180^o$ is an obtuse angle.

Hence, the answer is zero angle.

Find the supplement  of the following angle.
$40^{\circ}$

  1. $140$

  2. $40$

  3. $10$

  4. None of these


Correct Option: A
Explanation:

We know that the supplement angle

$=180^0-\theta$

So,
The supplement angle of $40^0$ will be
$=180^0-40^0=140^0$ 

Hence, this is the answer.

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